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$\def\d{\mathrm{d}}$Determine the integral $$\oint_L \mathbf{A} \cdot \,\d\mathbf{r},$$ where $$\mathbf{A} = \mathbf{e}_x(x^2-a(y+z))+\mathbf{e}_y(y^2-az)+\mathbf{e}_z(z^2-a(x+y)),$$ and $L$ is the curve given by the intersection between the cylinder $$\begin{cases}(x-a)^2+y^2=a^2 \\z\geq0\end{cases}$$ and the sphere $$x^2+y^2+z^2=R^2, \quad (R^2>4a^2)$$ The orientation is such that at $x=0$ the tangent to the curve is parallel with $-\mathbf{e}_y$.

Attempted solution:

Let's apply Stokes' theorem. First, let me introduce a graphical representation of the problem.enter image description here The path $L$ will then, as seen from above, be the following: enter image description here

A simple calculation shows $\nabla \times \textbf{A} = (0,0,a)$. Here comes my problem...

Question

What surface am I looking to take a surface integral over? Is it the whole cylinder or just the "top"? How can I determine?

Ѕᴀᴀᴅ
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Lozansky
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2 Answers2

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The surface can be ANY surface whose border is the curve $L$.

In this case, a possibility would be the part of the sphere inside the cylinder, which you can parametrize as follows: \begin{cases} x= x\\ y=y \quad \quad \quad \quad \quad \quad \text{with} \quad (x,y)\mid (x-a)^2+y^2\le a^2\\ z= \sqrt{R^2-x^2-y^2}\\ \end{cases}

Applying Stokes' theorem to finish the question yields \begin{align} \oint_L \textbf{A}\cdot d\textbf{r} = \iint_S \nabla\times \textbf{A}\cdot d\textbf{S} &= \iint_\limits{(x-a)^2+y^2\le a^2}\pmatrix{0\\0\\a}\cdot \pmatrix{1\\0\\\frac{-x}{\sqrt{R^2-x^2-y^2}}} \times \pmatrix{0\\1\\\frac{-y}{\sqrt{R^2-x^2-y^2}}}\; dA \\ &= \iint_\limits{(x-a)^2+y^2\le a^2}\pmatrix{0\\0\\a}\cdot \pmatrix{*\\*\\1}\; dA \\ &= \iint_\limits{(x-a)^2+y^2\le a^2}a\; dA = \pi a^3 \end{align}

Kuifje
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  • Should that limit be changed to $(x-a)^2+y^2 \leq a^2$ perhaps? And this area is the whole cylinder's surface area (including the bottom area) minus the area of the ellipse at the top? – Lozansky Apr 20 '17 at 08:38
  • @Lozansky Yes of course, you are right. I have edited. No, it is only the top part of the sphere inside the cylinder. You do not need to consider any ellipse whatsoever. – Kuifje Apr 20 '17 at 16:37
  • Alright, so it's the surface depicted in the last picture (enclosed by the vectors)? And if we "projected" that surface on the cylinder, it would be an ellipse? – Lozansky Apr 20 '17 at 17:22
  • @Lozansky Yes, it is the surface in the last picture. But no, the projection is a circle, and not an ellipse. In fact it is quite clear from the picture that it is a circle. The ellipse Vlad is referring to is the one in the first picture. You do not need it here. – Kuifje Apr 20 '17 at 18:14
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HINT: the intersection of sphere and cylinder is the curve $L$, which happened to be ab ellipse in this particular case since $R^2>4a^2$. Can you find the equation of the ellipse?

The interior of the ellipse is the surface you would use to compute integral on $L$ via Stoke's theorem.

Vlad
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  • Note that it doesn't have to be this ellipse. It can be any surface whose border is $L$. IMAO I don't think the ellipse is the best option here, as it requires unnecessary calculations for the question. – Kuifje Apr 20 '17 at 00:22
  • @Kuifje I know the choice of surface is not unique, so I tried to phrase my answer in a way that it would not create false impression uniqueness. On what surface do you think computing integral will be easier than on ellipse? Keep in mind that ellipse is flat and 2D so you can map it onto plane and integrate from there – Vlad Apr 20 '17 at 00:28
  • I think it's just simpler to use the part of the sphere inside the cylinder, because its equation is already known. – Kuifje Apr 20 '17 at 00:33
  • @Kuifje You would need to find equation of ellipse anyway, because the upper section of the sphere has the same boundary $L$ as flat ellipse. The difference is that in case of sphere you need to operate in 3D, whereas flat ellipse can be easily mapped to plane, so you would have one less dimension to worry about =) – Vlad Apr 20 '17 at 00:35
  • Sorry I disagree :) See my answer: you do not need the equation of the ellipse. In fact you barely need any calculations. – Kuifje Apr 20 '17 at 00:39